If $x^y=e^{x-y}, x> 0, $ then $\frac{dy}{dx}$ is equal to:

$\frac{log_ex}{(1+log_ex)^2}$

The correct answer is Option (2) → $\frac{log_ex}{(1+log_ex)^2}$

$x^y=e^{x-y}$

$y=\log_xe^{x-y}$

$=\frac{\log_ee^{x-y}}{\log_ex}=\frac{x-y}{\log x}$

$∴\frac{dy}{dx}=-\frac{dy}{dx}\log x-\frac{(x-y)}{x}×\frac{1}{(\log x)^2}$

$\frac{dy}{dx}(1+\log x)=\frac{y-x}{x(\log x)^2}$

$\frac{dy}{dx}=\frac{\log_ex}{(1+\log_ex)^2}$